Answer
$$\dfrac{4}{3}$$
Work Step by Step
$$Volume=V=\int_{0}^{1} \int_{x}^{2-x} (x^2+y^2) \ dy \ dx \\=\int_{0}^{1} (2x^2+\dfrac{(2-x)^3}{3}-\dfrac{7x^3}{3}] \ dx \\=\int_{0}^{1} [x^2y+\dfrac{y^3}{3}]_{x}^{2-x} \ dx \\=[ \dfrac{2x^3}{3}]_0^1+[\dfrac{(2-x)^4}{12}]_0^1-[\dfrac{7x^4}{12}]_0^1 \\ =\dfrac{2}{3}-\dfrac{1}{12}-\dfrac{1}{12}+\dfrac{(2)^4}{12} \\=\dfrac{4}{3}$$
